An Introduction to Mathematical Statistics and Its Applications (6th Edition)

Published by Pearson
ISBN 10: 0-13411-421-3
ISBN 13: 978-0-13411-421-7

Chapter 3 Random Variables - 3.4 Continuous Random Variables - Questions - Page 136: 9

Answer

$\color{blue}{F_Y(y) = \begin{cases} 0, & y\lt -1 \\ \dfrac{1}{2} + y +\dfrac{y^2}{2}, & -1\le y \lt 0 \\ \dfrac{1}{2}+y -\dfrac{y^2}{2}, & 0\le y \lt 1 \\ 1, & y \ge 1 \end{cases}}$

Work Step by Step

Rewrite $f_Y(y)$ as $\begin{align*} f_Y(y) &= \begin{cases} 0, & y\lt -1 \\ 1+y, & -1\le y\lt 0 \\ 1-y, & 0\le y\lt 1 \\ 0, & y\ge 1 \end{cases} \end{align*}$ Then, $\begin{align*} F_Y(y) &= P(Y\le y) \\ &= \begin{cases} \displaystyle\int_{-\infty}^y f_Y(u)\ du, & y\lt -1 \\ \displaystyle\int_{-\infty}^{-1} f_Y(u)\ du + \int_{-1}^y f_Y(u)\ du, & -1\le y \lt 0 \\ \displaystyle\int_{-\infty}^{-1} f_Y(u)\ du + \int_{-1}^0 f_Y(u)\ du + \int_0^y f_Y(u)\ du, & 0\le y \lt 1 \\ \displaystyle\int_{-\infty}^{-1} f_Y(u)\ du + \int_{-1}^0 f_Y(u)\ du + \int_0^1 f_Y(u)\ du + \int_1^y f_Y(u)\ du , & y \ge 1 \end{cases} \\ &= \begin{cases} \displaystyle\int_{-\infty}^y 0\ du, & y\lt -1 \\ \displaystyle\int_{-\infty}^{-1} 0\ du + \int_{-1}^y (1+u)\ du, & -1\le y \lt 0 \\ \displaystyle\int_{-\infty}^{-1} 0\ du + \int_{-1}^0 (1+u)\ du + \int_0^y (1-u)\ du, & 0\le y \lt 1 \\ \displaystyle\int_{-\infty}^{-1} 0\ du + \int_{-1}^0 (1+u)\ du + \int_0^1 (1-u)\ du + \int_1^y 0\ du , & y \ge 1 \end{cases} \\ &= \begin{cases} 0, & y\lt -1 \\ 0 + \left( u +\dfrac{u^2}{2}\right]_{-1}^y, & -1\le y \lt 0 \\ 0 +\left( u +\dfrac{u^2}{2}\right]_{-1}^0 +\left( u -\dfrac{u^2}{2}\right]_0^y, & 0\le y \lt 1 \\ 0 +\left( u +\dfrac{u^2}{2}\right]_{-1}^0 + \left( u +\dfrac{u^2}{2}\right]_0^1 + 0 , & y \ge 1 \end{cases} \\ &= \begin{cases} 0, & y\lt -1 \\ 0 + \left[\left( y +\dfrac{y^2}{2}\right) - \left( (-1) +\dfrac{(-1)^2}{2}\right)\right], & -1\le y \lt 0 \\ 0 +\left[0 - \left( (-1) +\dfrac{(-1)^2}{2}\right)\right] +\left[\left( y -\dfrac{y^2}{2}\right) - \left( -1 +\dfrac{1}{2}\right)\right], & 0\le y \lt 1 \\ 0 +\left[0 - \left( (-1) +\dfrac{(-1)^2}{2}\right)\right] +\left[\left( 1 -\dfrac{1^2}{2}\right) - 0\right] + 0 , & y \ge 1 \end{cases} \\ &= \begin{cases} 0, & y\lt -1 \\ y +\dfrac{y^2}{2} - \left( -1 +\dfrac{1}{2}\right), & -1\le y \lt 0 \\ -\left( -1 + \dfrac{1}{2}\right) +\left( y -\dfrac{y^2}{2}\right), & 0\le y \lt 1 \\ -\left( -1 +\dfrac{1}{2}\right) +\left( 1 -\dfrac{1}{2}\right), & y \ge 1 \end{cases} \\ &= \begin{cases} 0, & y\lt -1 \\ y +\dfrac{y^2}{2} +\dfrac{1}{2}, & -1\le y \lt 0 \\ \dfrac{1}{2}+y -\dfrac{y^2}{2}, & 0\le y \lt 1 \\ 1, & y \ge 1 \end{cases} \\ \color{blue}{F_Y(y)} &\color{blue}{= \begin{cases} 0, & y\lt -1 \\ \dfrac{1}{2} + y +\dfrac{y^2}{2}, & -1\le y \lt 0 \\ \dfrac{1}{2}+y -\dfrac{y^2}{2}, & 0\le y \lt 1 \\ 1, & y \ge 1 \end{cases}} \end{align*}$
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